Finding Voltage in a Circuit

3. Is there an easier way to do this question without applying mesh analysis, because this question was in my chapter 2 (intro to KVL/KCL), but i cannot think of a way to find this unless i use mesh analysis. Any advice

Well assuming I'm doing it right of course, you wouldn't need any simultaneous equations or anything, that'd be over complicating it, just need KCL really

so at the node above the 20 and 5 k resistors(or the equivalent 4k resistor!)you have 10 milliamps flowing in, and you need 10 mA flowing out, obtained with a voltage drop across that 4 k of just V=IR=10*4=40 Volts

You can check to see that puts 8 mA across the 5 k resistor and 2 across the 20 k, for a total of 10 mA

ok, so what you basically have are two resistors in series, and those two resistors(with eq. resistance of 4k) are in parallel with the other two resistors which I mentioned have an equiv. resistance of 4k as well

Since they're in parallel the current splits down both paths and equal amounts both ways since the resistances are ultimately equal, sooo it's actually V=I*R=5*4=20Volts